1) \(\sqrt[]{3x+7}-5< 0\)

\(\Leftrightarrow\sqrt[]{3x+7}< 5\)

\(\Leftrightarrow3x+7\ge0\cap3x+7< 25\)

\(\Leftrightarrow x\ge-\dfrac{7}{3}\cap x< 6\)

\(\Leftrightarrow-\dfrac{7}{3}\le x< 6\)

1.

ĐKXĐ: \(x< 5\)

\(\Leftrightarrow\sqrt{\dfrac{42}{5-x}}-3+\sqrt{\dfrac{60}{7-x}}-3=0\)

\(\Leftrightarrow\dfrac{\dfrac{42}{5-x}-9}{\sqrt{\dfrac{42}{5-x}}+3}+\dfrac{\dfrac{60}{7-x}-9}{\sqrt{\dfrac{60}{7-x}}+3}=0\)

\(\Leftrightarrow\dfrac{9x-3}{\left(5-x\right)\left(\sqrt{\dfrac{42}{5-x}}+3\right)}+\dfrac{9x-3}{\left(7-x\right)\left(\sqrt{\dfrac{60}{7-x}}+3\right)}=0\)

\(\Leftrightarrow\left(9x-3\right)\left(\dfrac{1}{\left(5-x\right)\left(\sqrt{\dfrac{42}{5-x}}+3\right)}+\dfrac{1}{\left(7-x\right)\left(\sqrt{\dfrac{60}{7-x}}+3\right)}\right)=0\)

\(\Leftrightarrow x=\dfrac{1}{3}\)

b.

ĐKXĐ: \(x\ge2\)

\(\sqrt{\left(x-2\right)\left(x-1\right)}+\sqrt{x+3}=\sqrt{x-2}+\sqrt{\left(x-1\right)\left(x+3\right)}\)

\(\Leftrightarrow\sqrt{\left(x-2\right)\left(x-1\right)}-\sqrt{x-2}+\sqrt{x+3}-\sqrt{\left(x-1\right)\left(x+3\right)}=0\)

\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x-1}-1\right)-\sqrt{x+3}\left(\sqrt{x-1}-1\right)=0\)

\(\Leftrightarrow\left(\sqrt{x-1}-1\right)\left(\sqrt{x-2}-\sqrt{x+3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{x-2}-\sqrt{x+3}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=1\\x-2=x+3\left(vn\right)\end{matrix}\right.\)

\(\Rightarrow x=2\)

a:ĐKXĐ: 3<=x<=5

Ta có: \(3x^2-17x+24=\sqrt{x-3}+3\sqrt{5-x}\)

=>\(3x^2-12x-5x+20=\sqrt{x-3}-1+3\sqrt{5-x}-3\)

=>\(\left(x-4\right)\left(3x-5\right)=\frac{x-3-1}{\sqrt{x-3}+1}+3\left(\sqrt{5-x}-1\right)\)

=>\(\left(x-4\right)\left(3x-5\right)=\frac{x-4}{\sqrt{x-3}+1}+3\cdot\frac{5-x-1}{\sqrt{5-x}+1}\)

=>\(\left(x-4\right)\left(3x-5-\frac{1}{\sqrt{x-3}+1}+\frac{3}{\sqrt{5-x}+1}\right)=0\)

=>x-4=0

=>x=4(nhận)

b: ĐKXĐ: x>=1

Ta có: \(\sqrt[3]{x+6}-2\sqrt{x-1}=4-x^2\)

=>\(\sqrt[3]{x+6}-2+2-2\cdot\sqrt{x-1}=4-x^2\)

=>\(\frac{x+6-8}{\sqrt[3]{\left(x+6\right)^2}+2\cdot\sqrt[3]{x+6}+4}+2\left(1-\sqrt{x-1}\right)+\left(x-2\right)\left(x+2\right)=0\)

=>\(\frac{x-2}{\sqrt[3]{\left(x+6\right)^2}+2\cdot\sqrt[3]{x+6}+4}+2\cdot\frac{1-x+1}{1+\sqrt{x-1}}+\left(x-2\right)\left(x+2\right)=0\)

=>\(\left(x-2\right)\left(\frac{1}{\sqrt[3]{\left(x+6\right)^2}+2\cdot\sqrt[3]{x+6}+4}+\frac{-2}{1+\sqrt{x-1}}+\left(x+2\right)\right)=0\)

=>x-2=0

=>x=2(nhận)

c: ĐKXĐ: \(x^3+3x^2+x-1\ge0\)

=>\(x^3+x^2+2x^2+2x-x-1\ge0\)

=>(x+1)\(\left(x^2+2x-1\right)\ge0\)

=>-1-\(\sqrt2\) <=x<=-1 hoặc \(x\ge-1+\sqrt2\)

\(x^2+5x+2=4\cdot\sqrt{x^3+3x^2+x-1}\)

=>\(x^2-x+6x-6=4\cdot\sqrt{x^3+3x^2+x-1}-8\)

=>(x-1)(x+6)=\(4\cdot\left(\sqrt{x^3+3x^2+x-1}-2\right)=4\cdot\frac{x^3+3x^2+x-1-4}{\sqrt{x^3+3x^2+x-1}+2}\)

=>(x-1)(x+6)=\(4\cdot\frac{x^3-x^2+4x^2-4x+5x-5}{\sqrt{x^3+3x^2+x-1}+2}\)

=>(x-1)(x+6)=4\(\frac{\left(x-1\right)\left(x^2+4x+5\right)}{\sqrt{x^3+3x^2+x-1}+2}\)

=>(x-1)\(\left\lbrack\frac{4\left(x^2+4x+5\right)}{\sqrt{x^3+3x^2+x-1}+2}-x-6\right\rbrack=0\)

=>x-1=0

=>x=1(nhận)

a: ĐKXĐ: -2<=x<=2

Đặt \(a=\sqrt{2+x};b=\sqrt{2-x}\)

Phương trình sẽ trở thành:

ab+6=2a+3b

=>ab-2a-3b+6=0

=>a(b-2)-3(b-2)=0

=>(b-2)(a-3)=0

=>b=2 hoặc a=3

=>2+x=9 hoặc 2-x=4

=>x=-7(loại) hoặc x=-2(nhận)

b: ĐKXĐ: x<=2

\(\left(\sqrt{2-x}+1\right)^2=3x+1\)

=>\(2-x+1+2\cdot\sqrt{2-x}=3x+1\)

=>\(-x+3+2\cdot\sqrt{2-x}-3x-1=0\)

=>\(2\cdot\sqrt{2-x}-4x+2=0\)

=>\(\sqrt{2-x}-2x+1=0\)

=>\(\sqrt{2-x}=2x-1\)

=>\(\begin{cases}2x-1\ge0\\ \left(2x-1\right)=\left(2-x\right)^2\end{cases}\Rightarrow\begin{cases}x\ge\frac12\\ x^2-4x+4=2x-1\end{cases}\)

=>\(\begin{cases}x\ge\frac12;x\le2\\ x^2-6x+5=0\end{cases}\)

=>(x-1)(x-5)=0 và 1/2<=x<=2

=>x=1

c:

ĐKXĐ: 6-5x>=0

=>5x<=6

=>x<=1,2

\(2\sqrt[3]{3x-2}-3\cdot\sqrt{6-5x}+16=0\)

=>\(2\cdot\sqrt[3]{3x-2}+4+12-3\cdot\sqrt{6-5x}=0\)

=>\(2\cdot\left(\sqrt[3]{3x-2}+2\right)+3\left(4-\sqrt{6-5x}\right)=0\)

=>\(2\cdot\frac{3x-2+8}{\sqrt[3]{\left(3x-2\right)^2}-2\cdot\sqrt[3]{3x-2}+4}+3\cdot\frac{16-6+5x}{4+\sqrt{6-5x}}=0\)

=>\(2\cdot\frac{3x+6}{\sqrt[3]{\left(3x-2\right)^2}-2\cdot\sqrt[3]{3x-2}+4}+3\cdot\frac{5x+10}{4+\sqrt{6-5x}}=0\)

=>\(\left(2\cdot\frac{3}{\sqrt[3]{\left(3x-2\right)^2}-2\cdot\sqrt[3]{3x-2}+4}+3\cdot\frac{5}{4+\sqrt{6-5x}}\right)\left(x+2\right)=0\)

=>x+2=0

=>x=-2(nhận)

d: ĐKXĐ: x>=1

\(\sqrt[3]{x+6}-2\cdot\sqrt{x-1}=4-x^2\)

=>\(\sqrt[3]{x+6}-2-2\cdot\sqrt{x-1}+2=4-x^2\)

=>\(\frac{x+6-8}{\sqrt[3]{\left(x+6\right)^2}+2\cdot\sqrt[3]{x+6}+4}+2\left(1-\sqrt{x-1}\right)=\left(2-x\right)\left(2+x\right)\)

=>\(\frac{x-2}{\sqrt[3]{\left(x+6\right)^2}+2\cdot\sqrt[3]{x+6}+4}+2\cdot\frac{1-x+1}{1+\sqrt{x-1}}=\left(2-x\right)\left(2+x\right)\)

=>\(\frac{x-2}{\sqrt[3]{\left(x+6\right)^2}+2\cdot\sqrt[3]{x+6}+4}-2\cdot\frac{x-2}{1+\sqrt{x-1}}-\left(2-x\right)\left(2+x\right)=0\)

=>\(\frac{x-2}{\sqrt[3]{\left(x+6\right)^2}+2\cdot\sqrt[3]{x+6}+4}-2\cdot\frac{x-2}{1+\sqrt{x-1}}+\left(x-2\right)\left(2+x\right)=0\)

=>\(\left(x-2\right)\left(\frac{1}{\sqrt[3]{\left(x+6\right)^2}+2\cdot\sqrt[3]{x+6}+4}-\frac{2}{1+\sqrt{x-1}}+\left(2+x\right)\right)=0\)

=>x-2=0

=>x=2(nhận)